# 62.Unique-Paths

## ้ข็ฎๅฐๅ

https://leetcode.com/problems/unique-paths/

https://www.jiuzhang.com/solutions/unique-paths

## ้ข็ฎๆ่ฟฐ

``````A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?

Example 1:
Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right

Example 2:
Input: m = 7, n = 3
Output: 28

Constraints:
1 <= m, n <= 100
It's guaranteed that the answer will be less than or equal to 2 * 10 ^ 9.``````

## ไปฃ็ 

### Approach #1 Recursion

``````class Solution {
public int uniquePaths(int m, int n) {
if (m == 1 || n == 1)        return 1;

return uniquePaths(m - 1, n) + uniquePaths(m , n - 1);
}
}``````

### Approach #2 Dynamic Programming

``````class Solution {
public int uniquePaths(int m, int n) {
int[][] dp = new int[m][n];
for (int[] arr: dp) {
Arrays.fill(arr, 1);
}

for (int row = 1; row < m; row ++) {
for (int col = 1; col < n; col++) {
dp[row][col] = dp[row - 1][col] + dp[row][col - 1];
}
}

return dp[m - 1][n - 1];
}
}``````

### #2

``````public class Solution {
public int uniquePaths(int m, int n) {
if (m == 0 || n == 0) return 1;

int[][] sum = new int[m][n];
for (int i = 0; i < m; i++) {
sum[i][0] = 1;
}
for (int i = 0; i < n; i++) {
sum[0][i] = 1;
}

for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
sum[i][j] = sum[i - 1][j] + sum[i][j - 1];
}
}
return sum[m - 1][n - 1];
}
}``````

### #3

``````public class Solution {
public int uniquePaths(int m, int n) {
if (m == 0 || n == 0) return 1;

int[][] sum = new int[m][n];
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (i == 0 || j == 0) {
sum[i][j] = 1;
} else {
sum[i][j] = sum[i - 1][j] + sum[i][j - 1];
}
}
}
return sum[m - 1][n - 1];
}
}``````

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